Question

The electronic potential V at a point on the circumference of a thin non- conducting disk of radius r and uniform charge density $\sigma$ is given by equation $V=4\sigma r$. Which of the following expression correctly represents electrostatic energy store in the electric field of a similar charge disk of radius R?

Answer 1

In the thin disc, consider a small ring elements of thickness dr located at a radial distance $r$ from centre

The differential area of the ring will be :-

$dA=4\pi rdr$

So, it contains a charge

$dq=\sigma dA=4\pi r\sigma dr$

Given that, potential in the ring, $V=4\sigma r$.

Thus, electrostatic energy stored in the ring :-

$dE=Vdq=\left(4\sigma r\right).\left(4\pi r\sigma dr\right)=16\pi {\sigma }^2{r}^{2}dr$

we get total energy by integrating it, $O\to R$.

$E=\int dE=16\pi {\sigma }^2{\int }_0^R{r}^{2}dr=16\pi {\sigma }^2{\left[\frac{r^3}{3}\right]}_0^R$

$E=\frac{16}{3}\pi {\sigma }^2{R}^3$